How to calculate the limit $\lim_{x \to 1} \frac{x + x^2 +\cdots+ x^n - n}{x - 1}$? How can I find the limit of the following function:
$$\begin{equation*}
\lim_{x \to 1}
\frac{x + x^2 +\cdots+ x^n - n}{x - 1}
\end{equation*}$$
Any help will be appreciated.
Thanks!
 A: Hint. Note that
$$\frac{x + x^2 +\dots+ x^n - n}{x - 1}=\frac{(x-1) + (x^2-1) +\dots+ (x^n - 1)}{x - 1}.$$
Moreover, for $k\geq 1$,
$$\lim_{x\to 1}\frac{x^k-1}{x-1}=\lim_{x\to 1}\left(x^{k-1}+x^{k-2}+\dots +x+1\right)=k.$$
A: Hint:
L'Hospital's rule works well here.
A: Remember that $$\sum_{i=1}^n x^i=\frac{x \left(x^n-1\right)}{x-1}$$ Let $x=y+1$ and consider 
$$S_n=\frac {-n+\sum_{i=1}^n x^i }{x-1}=\frac{\frac{(y+1) \left((y+1)^n-1\right)}{y}-n}{y}$$ Now, using the binomial theorem or Taylor series  $$(y+1)^n=1+n y+\frac{1}{2} (n-1) n y^2+\frac{1}{6} (n-2) (n-1) n y^3+O\left(y^4\right)$$
$$(y+1)^n-1=n y+\frac{1}{2} (n-1) n y^2+\frac{1}{6} (n-2) (n-1) n y^3+O\left(y^4\right)$$
$$\frac{(y+1) ^n-1}{y}=n +\frac{1}{2} (n-1) n y+\frac{1}{6} (n-2) (n-1) n y^2+O\left(y^3\right)$$
$$(y+1)\frac{(y+1) ^n-1}{y}=n+\frac{1}{2} n (n+1) y+\frac{1}{6} n \left(n^2-1\right) y^2+O\left(y^3\right)$$
$$(y+1)\frac{(y+1) ^n-1}{y}-n=\frac{1}{2} n (n+1) y+\frac{1}{6} n \left(n^2-1\right) y^2+O\left(y^3\right)$$ making 
$$S_n=\frac{1}{2} n (n+1) +\frac{1}{6} n \left(n^2-1\right) y+O\left(y^2\right)$$ which shows the limit and how it is approached when $y\to 0$.
A: The limit is the definition of the derivative of 
$$
f(x)=x+x^2+\dotsb+x^n
$$
at $x=1$. But 
$$
f'(x)=1+2x+\dotsb+nx^{n-1}
$$
and so
$$
f'(1)=1+2+\dotsb+n=\frac{n(n+1)}{2}.
$$
